InterviewSolution
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InterviewSolution
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Using the data (all values are in kilocalorie per mole at `25^(@)C`) given below, calculate the bond enegry of `C-C` and `C-H` bonds. `DeltaH^(Theta)` combustion of ethane `=- 372.0` `DeltaH^(Theta)` combustion of propane `=- 530.0` `DeltaH^(Theta)` for `C`( garphite) `rarr C(g) =+ 172.0` Bond enegry of `H-H` bond `=+ 104.0` `Delta_(f)H^(Theta) of H_(2)O(l) =- 68.0` `Delta_(f)H^(Theta) of CO_(2)(g) =- 94.0` |
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Answer» Correct Answer - `C - C = 82 k cal// C - H = 99 k cal` For formation of `C_(3)H_(8) 3C + 4H_(2) rarr C_(3)H_(8) Delta H_(1) = ?` For formation of `C_(2)H_(6) 2C + 3H_(2) rarr C_(2) H_(6), Delta H_(2) = ?` Thus (i) `Delta H_(1) = - [2 (C - C) + 8 (C - H)] + [3C_(s - g) + 4 (H - H)]` (ii) `Delta H_(2) = [1 (C - C) + 6 (C - H) ] + [2 C_(s - g) + 3 (H - H) ]` Let the bond energy of `C - C` and `C - H` bonds be x kcal and y kcal respectively. Then, we have (iii) `Delta H_(1) = - (2 x + 8y) + [3 xx 1724 xx 104]` and (iv) `Delta H_(2) = - (x + 6y) + [2 xx 172 + 3 xx 104]` Given (v) `C + O_(2) rarr CO_(2) , Delta H = - 94.0 kcal` (vi) `H_(2) + (1)/(2) O_(2) rarr CO , Detlta H = -68.0 kcal` (vii) `C_(2)H_(6) + (7)/(2)O_(2) rarr 2CO_(2) + 3H_(2) O , Delta H = - 372.0 kcal` (viii) `C_(2)H_(8) + 5O_(2) rarr 3 CO_(2) + 4H_(2) O, Delta H = - 530 kcal` (vii) we get (ix) (ix) `2 C + 3H_(2) rarr C_(2)H_(6), Delta H_(2) = - 20.0 kcal` Again `3-(v) + 4 xx (vi) (viii)` gives, (x) `3C + 4 H_(2) rarr C_(3)H_(8), Delta H_(1) = - 20.0 kcal` solving equation (iii), (iv), (ix) and (x) we get `x + 6y = 676, 2 x + 8y = 956` or `x = 82` kcal and `y = 99 kcal` hence bond energy of `C - C` bond = 82 kcal and bond energy of `C - H` bond = 99 kcal |
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